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stiks02 [169]
3 years ago
9

The following table lists the speed of sound in various materials. Use this table to answer the question.

Chemistry
2 answers:
defon3 years ago
5 0

Answer:

4.9sec

Explanation:

Lynna [10]3 years ago
4 0

Answer: Time taken by a sound impulse to travel through an aluminum rod 25 kilometers long is 4.9 seconds.

Solution:

Length of aluminium rod = 25 km = 25000 m (1 km=1000 m)

Time taken by sound to travel through aluminium = 5100 m/s (from the table given)

speed=\frac{distance}{time}=\frac{25000 m}{\text{time taken}}

time taken =\frac{distance}{speed}=\frac{25000 m}{5100 m/s}=4.9019 seconds\approx 4.9 seconds

Hence, the correct option is 4.9 seconds.

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Use kinetic molecular theory to explain why a gas takes the shape and volume of its container
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Answer:

Imo : Gas particles are in constant, random motion. The volume of gas particles is negligible in comparison to the volume of the container. There are no attractive forces between gas particles.

7 0
3 years ago
An iron chloride compound contains 55.85 grams of iron and 106.5 grams of chlorine. What is the most likely empirical formula fo
mart [117]

Answer:

FeCl_{3}

Explanation:

Moles =\frac {Given\ mass}{Molar\ mass}

mass of Fe = 55.85 g

Molar mass of Fe = 55.85 g/mol

<u>Moles of Fe = 55.85 / 55.85 = 1</u>

mass of Cl = 106.5 g

Molar mass of Cl = 35.5 g/mol

Moles of Cl = 106.5 / 35.5 = 3

Taking the simplest ratio for Fe and Cl as:

1 : 3

The empirical formula is = FeCl_{3}

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3 years ago
Be sure to answer all parts. Styrene is produced by catalytic dehydrogenation of ethylbenzene at high temperature in the presenc
svlad2 [7]

Answer:

a) ΔHºrxn = 116.3 kJ, ΔGºrxn = 82.8 kJ,  ΔSºrxn =  0.113 kJ/K

b) At 753.55 ºC or higher

c )ΔG =  1.8 x 10⁴ J

    K = 8.2 x 10⁻²

Explanation:

a)                                 C6H5−CH2CH3  ⇒  C6H5−CH=CH2  + H₂

ΔHf kJ/mol                    -12.5                           103.8                      0

ΔGºf kJ/K                        119.7                         202.5                      0

Sº J/mol                          255                          238                      130.6*

Note: This value was not given in our question, but is necessary and can be found in standard handbooks.

Using Hess law to calculate  ΔHºrxn we have

ΔHºrxn  = ΔHfº C6H5−CH=CH2 +  ΔHfº H₂ - ΔHºfC6H5−CH2CH3

ΔHºrxn =     103.8 kJ + 0 kJ  - (-12.5 kJ)

ΔHºrxn = 116.3 kJ

Similarly,

ΔGrxn = ΔGºf C6H5−CH=CH2 +  ΔGºfH₂ - ΔGºfC6H5CH2CH3

ΔGºrxn=   202.5 kJ + 0 kJ - 119.7 kJ  = 82.8 kJ

ΔSºrxn = 238 J/mol + 130.6 J/mol -255 J/K = 113.6 J/K = 0.113 kJ/K

b) The temperature at which the reaction is spontaneous or feasible occurs when ΔG becomes negative and using

ΔGrxn =  ΔHrxn -TΔS

we see that will happen when the term  TΔS  becomes greater than ΔHrxn since ΔS  is positive  , and so to sollve for T we will make ΔGrxn equal to zero and solve for T. Notice here we will make the assumption that  ΔºHrxn and ΔSºrxn remain constant at the higher temperature  and will equal the values previously calculated for them. Although this assumption is not entirely correct, it can be used.

0 = 116 kJ -T (0.113 kJ/K)

T = 1026.5 K  =  (1026.55 - 273 ) ºC = 753.55 ºC

c) Again we will use

                       ΔGrxn =  ΔHrxn -TΔS

to calculate ΔGrxn   with the assumption that ΔHº and ΔSºremain constant.

ΔG =  116.3 kJ - (600+273 K) x 0.113 kJ/K =  116.3 kJ - 873 K x 0.113 kJ/K

ΔG =  116.3 kJ - 98.6 kJ =  17.65 kJ = 1.8 x 10⁴ J ( Note the kJ are converted to J to necessary for the next part of the problem )

Now for solving for K, the equation to use is

ΔG = -RTlnK and solve for K

- ΔG / RT = lnK  ∴ K = exp (- ΔG / RT)

K = exp ( - 1.8 x 10⁴ J /( 8.314 J/K  x 873 K)) = 8.2 x 10⁻²

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there are 18 electrons in the outermost shell of argon atom

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